Question

Find three conescutive whole number whose sum os more than 45 but less than 54.

PLZ SOLVE THE EQUATION ​

Answer 1

Let the number is x, x+1, x+2

x+x+1+x+2 > 45

3x+3 > 45

3x > 42

x > 14

then

Ans is: 15,16,17

Answer 2

Let the three consecutive numbers be x, x+1 and x+2.

Now,

$\: \: \: \: \: \: \: x + x + 1 + x + 2 > 45 \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3x + 3 > 45 \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3x > 42 \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x > 14$

and

$\: \: \: \: \: \: \: \: x + x + 1 + x + 2 < 54 \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3x + 3 < 54 \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3x < 51 \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x < 17$

So,

$17 > x > 14$

CASE 1:-

If x=15, then the 3 numbers are 15, 16 and 17. Their sum = 48.

CASE 2:-

If x=16, then the 3 numbers are 16, 17 and 18. Their sum = 51.

But this is not possible because the sum should be less than 51.

So, the three consecutive whole numbers are 15, 16 and 17.

$\huge{\boxed{Answer: \: 15, \: 16 \: and \: 17}}$